If \(y=\tan ^{-1}\left(\frac{4 \sin 2 x}{\cos 2 x-6 \sin ^2 x}\right)\), then \(\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)\) at \(x=0\) is

\(\begin{aligned} & y=\tan ^{-1}\left(\frac{4 \sin 2 x}{\cos 2 x-6 \sin ^2 x}\right) \\ & =\tan ^{-1}\left[\frac{4(2 \sin x \cos x)}{\left(\cos ^2 x-\sin ^2 x\right)-6 \sin ^2 x}\right] \\ & =\tan ^{-1}\left(\frac{8 \sin x \cos x}{\cos ^2 x-7 \sin ^2 x}\right)\end{aligned}\)
\(=\tan ^{-1}\left(\frac{8 \tan x}{1-7 \tan ^2 x}\right) \)
\( =\tan ^{-1}\left(\frac{7 \tan x+\tan x}{1-7 \tan x \cdot \tan x}\right) \)
\( =\tan ^{-1}(7 \tan x)+\tan ^{-1}(\tan x) \)
\( \therefore \quad y=\tan ^{-1}(7 \tan x)+x \)
\( \therefore \quad \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{1}{1+(7 \tan x)^2} \cdot 7 \sec ^2 x+1\)
\(=\frac{7 \sec ^2 x}{1+49 \tan ^2 x}+1 \)
\( \therefore \quad\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{x=0}= \)
\( =\frac{7 \sec ^2 0}{1+49 \tan ^2 0}+1 \)
\( =\frac{7}{1+0}+1 \)
\( =8\)